`a)` 
$\widehat{B} = 90^\circ - \widehat{C} = 90^\circ - 30^\circ = 60^\circ$.
$\tan C = \frac{AB}{AC}$
$\Leftrightarrow \tan 30^\circ = \frac{AB}{100}$
$\Leftrightarrow AB = 100 \cdot \tan 30^\circ = 100 \cdot \frac{\sqrt{3}}{3} = \frac{100\sqrt{3}}{3}$ cm.
$\cos C = \frac{AC}{BC}$
$\Leftrightarrow \cos 30^\circ = \frac{100}{BC}$
$\Leftrightarrow BC = \frac{100}{\cos 30^\circ} = \frac{100}{\frac{\sqrt{3}}{2}} = \frac{200}{\sqrt{3}} = \frac{200\sqrt{3}}{3}$ cm.

`b)` 
$\widehat{B} = 90^\circ - \widehat{C} = 90^\circ - 45^\circ = 45^\circ$.
Do $\widehat{B} = \widehat{C} = 45^\circ$, $\triangle ABC$ là tam giác vuông cân tại $A$.
$\Rightarrow AC = AB = 50$ cm.
Áp dụng định lý Pitago:
$BC^2 = AB^2 + AC^2 = 50^2 + 50^2 = 2 \cdot 50^2$
$\Leftrightarrow BC = \sqrt{2 \cdot 50^2} = 50\sqrt{2}$ cm.

c) 
$\widehat{C} = 90^\circ - \widehat{B} = 90^\circ - 35^\circ = 55^\circ$.
$\sin B = \frac{AC}{BC}$
$\Leftrightarrow \sin 35^\circ = \frac{AC}{40}$
$\Leftrightarrow AC = 40 \cdot \sin 35^\circ \approx 40 \cdot 0.5736 \approx 22.944$ cm.
$\cos B = \frac{AB}{BC}$
$\Leftrightarrow \cos 35^\circ = \frac{AB}{40}$
$\Leftrightarrow AB = 40 \cdot \cos 35^\circ \approx 40 \cdot 0.8192 \approx 32.768$ cm.

`d)` 
Áp dụng định lý Pitago:
$BC^2 = AB^2 + AC^2 = 70^2 + 60^2 = 4900 + 3600 = 8500$
$\Leftrightarrow BC = \sqrt{8500} = \sqrt{2500 \cdot 3.4} = 50\sqrt{3.4}$ cm.
$\tan B = \frac{AC}{AB} = \frac{60}{70} = \frac{6}{7}$
$\Leftrightarrow \widehat{B} = \arctan(\frac{6}{7}) \approx 40.6^\circ$.
$\widehat{C} = 90^\circ - \widehat{B} \approx 90^\circ - 40.6^\circ = 49.4^\circ$.